What is the solution for y^{''}+k^2y=acos(bx),y(0)=0,y^'(0)=0 ?

The solution for y^{''}+k^2y=acos(bx),y(0)=0,y^'(0)=0 is y=(acos(bx))/(k^2-b^2)-(acos(kx))/(k^2-b^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+k^2y=acos(bx),y(0)=0,y^'(0)=0

y^{^{''}} + k^{2} y = a cos (b x), y (0) = 0, y^{^{'}} (0) = 0

y = \frac{a cos ( b x )}{k ^{2} - b ^{2}} - \frac{a cos ( k x )}{k ^{2} - b ^{2}}

Solution steps

y^{''} (x) + k^{2} y = a cos (b x)

Solve linear ODE:

y = \frac{a cos ( b x )}{k ^{2} - b ^{2}} - \frac{a cos ( k x )}{k ^{2} - b ^{2}}

y = \frac{a cos ( b x )}{k ^{2} - b ^{2}} - \frac{a cos ( k x )}{k ^{2} - b ^{2}}

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{x} (\frac{1}{sin ( x )})

y^{^{''}} - y^{^{'}} - 6 y = 5 t, y (0) = 3, y^{^{'}} (0) = 1

D, D (x) = x^{8} - 4^{4}

d (x) = 120 - 4 x^{2}

x^{^{''}} + 2 x^{^{'}} + 3 x = t^{2} + 4 t

What is the solution for y^{''}+k^2y=acos(bx),y(0)=0,y^'(0)=0 ?
The solution for y^{''}+k^2y=acos(bx),y(0)=0,y^'(0)=0 is y=(acos(bx))/(k^2-b^2)-(acos(kx))/(k^2-b^2)